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Mar 25, 2018 at 16:56 comment added Piotr Achinger Yes, that's how I understood your question. Let me explain a bit more: start with $V,K,k$ as in your question. There is a natural finite map $W(k)\to V$ identifying $V$ with the integral closure of $W(k)$ in $K$. It follows that the integral closures of $W(k)$ and $V$ in $\bar K$ coincide, and so $W(k)$ and $K$ yield the same $A_N$.
Mar 25, 2018 at 14:08 comment added Konstantin @PiotrAchinger I think we are misunderstanding each other: I would like to find out what the above machinery gives me when I put in $W(k)=V$ then $K=Frac(W(k))$ etc. So I am actually looking for the integral closure of $W(k)$ in $\overline{Frac(W(k))}$ not $\overline{Frac(V)}$
Mar 25, 2018 at 13:41 comment added Piotr Achinger In any case Serre's "Local Fields" should be sufficient.
Mar 25, 2018 at 13:40 comment added Piotr Achinger This is just transitivity of integral closure: you have $W(k)\subseteq K\subseteq \bar K$ and $V$ is the integral closure of $W(k)$ in $K$.
Mar 25, 2018 at 12:47 comment added Konstantin @PiotrAchinger Thank you very much. I know that the construction only depends on $\overline{K}$, what I dont know is that $V$ $W(k)$ have the same integral closure in $\overline{K}$. I am sure this is very basic but my knowldege on Witt rings comes mainly (solely) from Serre's "Local Fields" which is probably not enough at this point. Could you give me a refeference for a result like this?
Mar 25, 2018 at 12:04 comment added Piotr Achinger The construction depends only on $\bar K$, so $V$ and $W(k)\subseteq V$ give the same result as their integral closures in $\bar K$ are the same (they both consist of all elements of non-negative valuation).
Mar 25, 2018 at 11:33 history asked Konstantin CC BY-SA 3.0